SudoCue suggests 3D Medusa Coloring with 2 linked clusters that eliminates (3)r5c6. One of the clusters (a long one) ends at (6)r45c6, the other shown by SudoCue is a short one, with only 2 cells: (3)r4c9=(3)r5c7.

Above the colors of the chains are marked with */# and @/$. I don't see the link between the two chains. I tried adding (3)r4c6 to the short chain thus making a bridge between the two but it doesn't make a conflict for (3)r5c6 (the colors are not opposite).

I could make a conflict with a third cluster though: (4)r1c2 - (4)r1c9=(4)r8c9=(4=5)r9c8=(5)r7c9=(5)r4c9 - (3)r4c9 (I wrote the 2 weak links making the bridges also to the ends of this chain -- (4)r1c2 also belongs to the big first chain with the same color as (6)r4c6). Is this the Medusa SudoCue suggests (maybe, as SudoCue shows two yellow cells as well: (4)r1c9 and (5)r4c9 shown with ! above) or is there a simpler bridge?

I can see a little bit simpler pattern making the same elimination: If r1c3 couldn't be 3 then r1c69 with box 6 would be an Empty Rectangle on candidate 3 eliminating r5c6. Now if r1c3==3 then there is a short chain: r7c3==1 => r4c3==6 => r4c6<>6 => r5c6==6, so (3)r5c6 can be removed.

Well, this combination of an ER and a 3D Chain is not easy to spot but probably easier than 3 bridged clusters. But of course they are not equivalent...

I have to say I agree with you, but I must admit I'm still mostly a 2-D thinker, as I haven't really begun trying to use Medusa coloring as such. I could not reproduce the clusters that you posted, even after playing around with my solver settings. When I finally did come up with a Medusa coloring position, it was a different one than what you showed. Maybe we have different versions of the Sudocue solver. Anyway, taking you at your word, I did not see any obvious way to join up the two chains you show. It doesn't seem like a chain with only two cells can be of much use.

I'm beginning to feel comfortable with grouped chains, and I think that in some cases they can in effect bridge the gap between two Medusa clusters. For what it's worth, in my solution I found a grouped chain in a position somewhat like yours. It still holds in your position (reproduced below).

Here, since digit "8" is locked into all sides of the "38" UR pattern in r49c45 (marked with "*"), (3)r4c45 can be eliminated. After the fact I found that the Sudocue solver does not choose this path, evidently due to the order in which the various techniques are attempted. I was stuck quite a while after this elimination, but eventually found the AIC which I mentioned in my first post:

What I missed -- the Sudocue solver finds the ALS XZ rule elimination (if this technique is enabled) represented by this AIC:

(6=4)r1c2 - (4=23896)r1c9|r2c789|r3c7 => r1c8, r2c3 <> 6

Either AIC enables us to place all remaining "6's" in the grid.

If the ALS XZ rule technique is disabled, then the Sudocue solver finds some bridged Medusa cluster eliminations. So far I haven't been able to understand them, but I should spend more time trying.

After the elimination of (6)r5c2 above, I continued with basic techniques, a finned X-wing eliminating (2)r7c9, a short XY chain eliminating (5)r4c9 and (5)r9c8, and a naked "123" triple in r4c169, to reach this position:

There are other ways to proceed, but the threatening 35-58-83 deadly pattern in r249c45 (marked with "*") stands out here. The deadly pattern can only be avoided by placing "9" in one of r2c45, so that (9)r2c39 and (9)r3c5 can be eliminated.

As an aside, I did consider other patterns here but none really seemed to add anything more. For example, in cells r6c56, r7c6|r8c5, and r78c7 (marked with "#") there is a 6-cell pattern based on "79." Digit "7" is locked into all sides of the pattern, so to avoid the deadly pattern, either the "9" in box 5 or box 9 must lie outside of the pattern. This means either r6c6=7 or r7c7=7, so that in either case r7c6 <> 7. This result, however, follows quickly from the elimination of (9)r2c9 given by the first pattern (which leaves r2c8 with a hidden single for "9" in column 9).

Actually, there wasn't much more needed anyway, since this is the position after basic follow-up:

It's a bit of overkill, but we can use a BUG+2 argument to complete the solution. Since this isn't yet addressed in the Solving Guide, I'll expound a bit. First, to quote the Solving Guide, a BUG is "a situation in which every unsolved cell has 2 candidates and every row, column and box have 2 candidates for each unsolved digit." As also stated in the Solving Guide, a BUG position has either 0 or 2 solutions. In the diagrammed position, there are only two unsolved cells -- r3c3 and r3c8 -- which have more than two candidates (thus the term BUG+2), and if both of the surplus candidates (3)r3c3 and (9)r3c8 were removed, a BUG would result. Any placement of a candidate which would lead to elimination of both of these surplus candidates must be invalid, and it's easy to see that setting r3c5 to "3" would do just that, as r3c7 would then become "9."

I suppose we could express this in an AIC, something like:

(3)r3c3 = BUG+2 = (9)r3c8 - (9=3)r3c7 => r3c5 <> 3

In this case, an XY wing which gives the same conclusion is easy to spot: